Question

From the list given below, the number of lanthanides which exhibit +4 state in their compounds is: Given lanthanides: Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

Key points about lanthanide oxidation states: 1. +3 is the most common state for all lanthanides
2. +4 state occurs when it creates special stability (empty/half-filled/filled f-subshell)
3. +2 state appears when it leads to f$^7$ or f$^{14}$ configurations
4. Ce$^{4+}$ (not in list) is the most stable +4 lanthanide ion

Answer 1

The Correct Option is B

Lanthanides exhibiting +4 state: - Praseodymium (Pr): Exhibits +4 state (4f$^1$ configuration).
- Terbium (Tb): Shows +4 state (4f$^7$ half-filled configuration).
- Neodymium (Nd): Can achieve +4 state (4f$^2$ configuration). Lanthanides not showing +4 state: - Promethium (Pm), Samarium (Sm), Europium (Eu), Gadolinium (Gd), Dysprosium (Dy). Thus, 3 lanthanides (Pr, Nd, Tb) exhibit +4 oxidation state, making option (2) correct.

Answer 2

Step 1: Understand oxidation states of lanthanides.
Lanthanides typically exhibit a +3 oxidation state, but some elements can also show +4 or +2 states depending on their electronic configurations.

Step 2: Identify lanthanides with +4 oxidation state.
Among the lanthanides, only a few exhibit the +4 state commonly in their compounds:
- Praseodymium (Pr) can show +4.
- Terbium (Tb) is well known for +4 state.
- Cerium (Ce) (not listed here) also exhibits +4.

From the given list:
Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy,
the elements exhibiting +4 oxidation state are:
- Pr (Praseodymium),
- Tb (Terbium),
- Possibly one more depending on conditions, but generally only these two are well-known.

However, sometimes Neodymium (Nd) and Dysprosium (Dy) can show +4 in very specific compounds, but it is rare.
The widely accepted count of lanthanides from the list showing +4 state is 3.

Step 3: Conclusion.
Therefore, the number of lanthanides exhibiting +4 state among the given is 3.